Electrical Engineering - Parallel Circuits - Discussion
Discussion Forum : Parallel Circuits - General Questions (Q.No. 1)
1.
An ammeter has an internal resistance of 50 
. The meter movement itself can handle up to 1 mA. If 10 mA is applied to the meter, the shunt resistor, RSH1, is approximately
. The meter movement itself can handle up to 1 mA. If 10 mA is applied to the meter, the shunt resistor, RSH1, is approximately55
5.5
50
9
Discussion:
16 comments Page 1 of 2.
Arjun said:
6 years ago
Rsh = Rm/(m-1).
Where as Rm = Meter Resistance.
m = Multiplying Factor ( 10mA/1mA = 10).
Rsh = 50/(10-1) = 5.5 ohm.
Where as Rm = Meter Resistance.
m = Multiplying Factor ( 10mA/1mA = 10).
Rsh = 50/(10-1) = 5.5 ohm.
(10)
Mq.Khan said:
4 years ago
Both current and resistance is known for 50 ohms.
Hence;
Voltage = 50 x 1 mA = 0.05.
Voltage is same in parallel.
Hence for unkown resistor
R= V/I.
R= 0.05/9 mA = 5.5 ohm.
Hence;
Voltage = 50 x 1 mA = 0.05.
Voltage is same in parallel.
Hence for unkown resistor
R= V/I.
R= 0.05/9 mA = 5.5 ohm.
(6)
Sumant said:
1 decade ago
Supplying current = Is = 10mA.
Ammeter current = Ia = 1mA.
The circuit is in parallel, So it is current divider circuit.
Therefore,
Ish = Is-Ia = 10mA-1mA = 9mA.
Va = Vsh = Vs = Ra*Ia = 50*1*10^-3 = 50mV.
Rsh = Vsh/Ish = (50*10^-3)/(9*10^-3) = 5.55V.
Answer B is correct.
Ammeter current = Ia = 1mA.
The circuit is in parallel, So it is current divider circuit.
Therefore,
Ish = Is-Ia = 10mA-1mA = 9mA.
Va = Vsh = Vs = Ra*Ia = 50*1*10^-3 = 50mV.
Rsh = Vsh/Ish = (50*10^-3)/(9*10^-3) = 5.55V.
Answer B is correct.
(2)
Harris said:
7 years ago
There is a much simpler formula to solve this.
R(sh)= Ig/(It-ig) * Rg.
Ig= current through meter,
It= Total Current,
Rg= Meter Resistance,
And we get; (0.001/0.009) * 50 = 5.5 ohms.
R(sh)= Ig/(It-ig) * Rg.
Ig= current through meter,
It= Total Current,
Rg= Meter Resistance,
And we get; (0.001/0.009) * 50 = 5.5 ohms.
(2)
Hemanth said:
6 years ago
The resistance is calculated by voltage equtation.
R = 50*1/1000*1000/9 = 5.5ohm.
R = 50*1/1000*1000/9 = 5.5ohm.
(2)
Anand said:
1 decade ago
Can anybody explain the answer ?
Bharamu said:
1 decade ago
Hi Anand, here is the answer.
See the meter resistance is 50 ohm, it carries max of 1mA current. Now if we wanna measure 10mA with same meter then we have to connect one low resistance across it so that we can bypass extra 9mA current through it.
So formula is
Rsh = Rm/(m-1)
Where as Rm = Meter Resistance
m = Multiplying Factor ( 10mA/1mA = 10)
Rsh = 50/(10-1) = 5.5 ohm
See the meter resistance is 50 ohm, it carries max of 1mA current. Now if we wanna measure 10mA with same meter then we have to connect one low resistance across it so that we can bypass extra 9mA current through it.
So formula is
Rsh = Rm/(m-1)
Where as Rm = Meter Resistance
m = Multiplying Factor ( 10mA/1mA = 10)
Rsh = 50/(10-1) = 5.5 ohm
Muna said:
1 decade ago
The resistance is calculated by voltage equtation.
R = 50*1/1000*1000/9 = 5.5ohm.
R = 50*1/1000*1000/9 = 5.5ohm.
Dushyant Dutta said:
1 decade ago
Rg*Ig/I-Ig=S
Raviraj said:
1 decade ago
See is the meter resistance is ohm.it Carries max of 1mA current.Nowif wb Wanne mesure 10mA With Same meter then we have to connect one low resistance across it so that we can Bypass axtra 9mA current through it
So Formula is Res=RM/(m-1)
Where as RM=Meter Resistance
m=Multi factor(10mA/1mA=10)
Res=50/(10-1)=5.5ohm Ans This one.
So Formula is Res=RM/(m-1)
Where as RM=Meter Resistance
m=Multi factor(10mA/1mA=10)
Res=50/(10-1)=5.5ohm Ans This one.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers